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Question

Let A=Q×Q, where Q is the set of all rational numbers, and * be a binary operation on A defined by (a, b) * (c, d) = (ac, b + ad) for (a.b),(c,d)ϵA. Then find

(i) The identity element of * in A.

(ii) Invertible elements of A, and hence write the inverse of elements (5, 3) and. (12,4).

OR

Let f : WW be defined as

f(n){n1,if n is oddn+1,if n is even
Show that f is invertible and find the inverse of f. Here, W is the set of all whole numbers.


Solution

Let A=Q×Q, where Q is the set of rational numbers.

Given that * is the binary operation on A defined by (a, b) * (c, d) = (ac, b + ad) for

(a,b),(c,d)^iA.

(i)

We need to find the identity element of the operation * in A.

Let (x, y) be the identity element in A.

Thus,

(a, b) * (x, y) = (x, y) * (a, b) = (a, b), for all (a,b)ϵA

(ax,b+ay)=(a,b)

ax=a and b+ay=b

y=0 and x=1

Therefore, (1,0)ϵA is the identity element in A with respect to the operation *.

(ii)

We need to find the invertible elements of A.

Let (p, q) be the inverse of the element (a, b)

Thus,

(a,b)(p,q)=(1,0)

(ap,b+aq)=(1,0)

ap=1 and b+aq=0

p=1aand q=ba

Thus the inverse elements of (a, b) is (1a,ba)

Now let us find the inverse of (5, 3) and (12,4)

Hence, inverse of (5, 3) is (12,35)

And inverse of  (12,4)is(2,412)=(2,8)

OR

Let f: WW be defined as

f(n){n1,if n is oddn+1,if n is even

We need to prove that 'f' is invertible.

In order to prove that 'f' is invertible it is sufficient to prove that f is a bijection.

A function f: AB is a one-one function or an injection, if

f(x)=f(y)x=y for all x,yϵA.

Case i:

If x and y are odd.

Let f(x) = f(y)

x1=y1

x=y

Case ii:

If x and y are even,

Let f(x) = f(y)

x+1=y+1

x=y

Thus, in both the cases, we have,

f(x)=f(y)x=y for all x,yϵW.

Hence f is an injection.

Let n be an arbitrary element of W.

If n is an odd whole number, there exists an even whole number n1ϵW such that

f(n - 1) = n − 1 + 1 = n.

If n is an even whole number, then there exists an odd whole number n+1ϵW

such that f(n + 1) = n + 1 − 1 = n.

Also, f(1) = 0 and f(0) = 1

Thus, every element of W (co-domain) has its pre-image in W (domain).

So f is an onto function.

Thus, it is proved that f is an invertible function.

Thus, a function g:BA which associates each element yϵB to a unique element xϵA

such that f(x) = y is called the inverse of f.

That is, f(x)=yg(y)=x

The inverse of f is generally denoted by f1.

Now let us find the inverse of f.

Let x,yϵW such that f(x) = y

x+1=y, if x is even

And

x − 1 = y, if x is odd

x={y1,if y is oddy+1,if y is even

f1(y)={y1,if y is oddy+1,if y is even

Interchange, x and y, we have,

f1(x)={x1,if x is oddx+1,if x is even144

Rewriting the above we have,

f1(x)={x1,if x is evenx+1,if x is oddThus,f1(x)=f(x)

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